Question

How many geometrical isomers are possible in the following two alkenes?
(i) $C{H}_3-CH=CH-CH=CH-C{H}_3$
(ii)$C{H}_3-CH=CH-CH=CH-Cl$

• A. 4 and 4
• B. 4 and 3
• C. 3 and 3
• D. 3 and 4

Answer 1

The correct option is D 3 and 4

1. Compound (i) is a symmetrical compound. It has same methyl terminal groups.
The formula used to determinne the number of geometrical isomers in compound having same terminal groups with even number of double is, $2^{(n-1)}+2^{\left[\right(n/2)-1]}$
where, n is the number of double bond present in the molecule.
In the given molecule we have two double bond so,
No. of geometrical isomers = $2^{(n-1)}+2^{\left[\right(n/2)-1]}$
= $2^{(2-1)}+2^{\left[\right(2/2)-1]}$
=$2^1+2^0$
=2+1
=3

2. Compound (ii) is a unsymmetrical compound. It has different terminal groups (methyl and chlorine).
The formula used to determinne the number of geometrical isomers in compound having different terminal groups is, $2^n$, where n is the number of dounle bond present in the molecule.
In the given compound we have 2 double bond so n = 2
Therefore,
No. of geometrical isomers = $2^n$
= $2^2$
= 4.
Therefore, the correct option is (d).